# multiphase interfoam k-omega BC

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 October 7, 2020, 23:58 multiphase interfoam k-omega BC #1 New Member   Diego Villegas Díaz Join Date: Oct 2020 Posts: 4 Rep Power: 4 Hi, I am a new foamer. I am simulating a section of an openchanel steady water flow with interFoam solver. The inlet condition is a water profile and a constant flow in. To represent this situation I used this boundary condition: 1) alpha.water: I used funkySetBoundaryFields to map the waterprofile to alpha.water variable. 1 for water and 0 for air, without intermediate values. 2) U: I used the variableHeightFlowRateInletVelocity with the inlet flow. 3) p_rgh: zero gradient. 4) nut: calulated. 5) k: turbulentIntensityKineticEnergyInlet; with Intensity = 0.0573 6) omega: turbulentMixingLengthFrequencyInlet; with mixing length = 0.00258 I've had some troubles with the k and omega boundary conditions. interFoam fall at the start and I dont understand why. Meanwhile, I am using a fixed value for omega (3.46) or fixed value for k (0.0044065), and its work but I really want to understand why when I use the boundary condition 5) and 6) together interFoam fall. Thaks in advance.

 October 8, 2020, 02:08 #2 Senior Member   Charles Join Date: Aug 2016 Location: Vancouver, Canada Posts: 138 Rep Power: 8 I'm curious how do you get these values, i.e. intensity 0.0573, mixing length 000258, omega 3.46 and k 0.0044065. I guess the divergence was caused by improper initial conditions because turbulence model is prone to diverge.

October 8, 2020, 09:53
#3
New Member

Diego Villegas Díaz
Join Date: Oct 2020
Posts: 4
Rep Power: 4
Quote:
 Originally Posted by Marpole I'm curious how do you get these values, i.e. intensity 0.0573, mixing length 000258, omega 3.46 and k 0.0044065. I guess the divergence was caused by improper initial conditions because turbulence model is prone to diverge.
I get these values using the water profile at the inlet and a median velocity (U) . I calculated the intensity using some expresion in the book
"Turbulence in open channel flows" (Nezu & Nakagawa) the mixing length scale, k = (3/2)*(I*U)^2, the mixing length for the omega condition I calculated as 0.07*Rh (with Rh the hydraulic radius) I found this scale in these forum but I dont have any formal information. Finally to calculate omega (for fixed value), I calculated another length scale to mixing length lm equal to 0.41*Rh (based in Von Karman constant) and omega is equal to sqrt(k)/lm.

I forgot to mention that my system is initialized at the inlet as uniforms values for k, omega and lm, the same values that I mentioned above. May be these unifor initial condition is the reason why interFoam falls.

 October 8, 2020, 21:10 #4 Senior Member   Charles Join Date: Aug 2016 Location: Vancouver, Canada Posts: 138 Rep Power: 8 k = (3/2)*(I*U)^2 is not mixing length, but is turbulent kinetic energy. So, I guess you need to review these initial values.

October 8, 2020, 23:41
#5
New Member

Diego Villegas Díaz
Join Date: Oct 2020
Posts: 4
Rep Power: 4
Quote:
 Originally Posted by Marpole k = (3/2)*(I*U)^2 is not mixing length, but is turbulent kinetic energy. So, I guess you need to review these initial values.
Ohh I made a mistake writing the last post. I know that k is turbulent Kinect energy. But I'm thinking that interFoam fall due the uniform initial condition un k and omega.

 October 9, 2020, 02:44 #6 Senior Member   Charles Join Date: Aug 2016 Location: Vancouver, Canada Posts: 138 Rep Power: 8 You have two mixing length. The first mixing length 0.07Rh might be problematic and is too small. You can try 0.41Rh instead. DiegoAlonso likes this.

 Tags boundary condition, interfoam, k-omega model, open channel flow